How do you verify #(sinA+cosA)^2=(2+secAcscA)/(secAcscA)#?

1 Answer
Bdub
Nov 12, 2016

see below

Explanation:

#(sinA+cosA)^2=(2+secAcscA)/(secAcscA)#

#(sinA+cosA)(sinA+cosA)=(2+1/cosA*1/sinA)/(1/cosA*1/sinA)#

#sin^2A+2sinAcosA+cos^2A=((2sinAcosA+1)/(cosAsinA))/(1/(cosAsinA))#

#sin^2A+2sinAcosA+cos^2A=(2sinAcosA+1)/(cosAsinA)*(cosAsinA)/1#

#2sinAcosA+sin^2A+cos^2A=(2sinAcosA+1)/cancel(cosAsinA)*cancel(cosAsinA)/1#

#2sinAcosA+1=2sinAcosA+1#

#:.#Left Hand Side=Right Hand Side

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